please tell me how to find the x coordinate of the absolute maximum and minimum of the function 9x+(7/x), x>0

Answer 1

Take the derivative of the function and set it to zero. Then solve for x.

You have 9x + 7x^-1

The derivative is 9 + (-1)7x^-2

That's 9 - 7/(x^2) which we set to 0.

9 - 7/x^2 = 0
9 = 7/x^2
x^2 = 7/9
x = sqrt(7/9)
x = sqrt(7)/3

x = +/- 0.881917104

Your problem statement says x > 0, so that means we only care about x = +0.881917104. Looking at the graph given below (or taking the second derivative) you can confirm that this is the absolute minimum of the positive portion of the graph.

There is no absolute maximum because the graph asymtotically approaches the y-axis (x=0) and also grows forever in the positive direction thus it will have no maximum.

Answer 2

to find max or min u find first derivative wrt x

so if f(x) = 9x +(7/x)
then d(f(x))/dx = 9- 7/(x^2)

to find out max or min you put it equal to zero:

9 - 7/(x^2) = 0
or x^2 = 7/9
or x = sqrt(7/9) and x = -sqrt(7/9)

so these are the two values of x-oordinates
now to find out which one is max and which is min, find double derivative of f(x)

d2(f(x)/(dx2) = 14/x^3
so naturally the point at which this is <0, the point is max.(inthis case x = sqrt(7/9)
and thepoint where this double derivative is >0, it is min, in this case x = -sqrt(7/9)